Question

Let
A=2x^2i−3yzj+xz^2k
and
ϕ=2z−x^3y
, find
A.▽ϕ
at point (1,-1,1).
5
3
4
1

Answer 1

Answer on Question #56127 – Math – Vector Calculus

Let

A=2x^2i-3yzj+xz^2k

and

φ=2z-x^3y

, find

A. ∇φ

at point (1,-1,1).

5

3

4

1

Solution

$$\nabla \varphi = \langle \frac{\partial}{\partial x} (2z - x^3y), \frac{\partial}{\partial y} (2z - x^3y), \frac{\partial}{\partial z} (2z - x^3y) \rangle = \langle -3x^2y, -x^3, 2 \rangle,$$ $$A = \langle 2x^2, -3yz, xz^2 \rangle$$ $$\left. \nabla \varphi \right|_{(1,-1,1)} = \langle -3 \cdot 1^2(-1), -1^3, 2 \rangle = \langle 3, -1, 2 \rangle,$$ $$\left. A \right|_{(1,-1,1)} = \langle 2 \cdot 1^2, -3 \cdot (-1) \cdot 1, 1 \cdot 1^2 \rangle = \langle 2, 3, 1 \rangle$$ $$A \cdot \left. \nabla \varphi \right|_{(1,-1,1)} = 3 \cdot 2 + (-1) \cdot 3 + 2 \cdot 1 = 5$$

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